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1 Serial :. T_CS_A_Operating System_ Delhi Noida Bhopal yderabad Jaipur Lucknow Indore une Bhubaneswar Kolkata atna Web: h: - CLASS TEST - COMUTER SCIENCE & IT Subject : Operating Systems Date of test : // Answer Key. (b). (b). (c). (d). (b). (c). (d). (d). (b). (b). (d). (c). (b). (c). (c). (c). (b). (a). (a). (c). (c). (d). (c). (c). (a). (a). (b). (c). (c). (b)
2 CT- CS Operating Systems Detailed Explanations. (b) If n process completing with waiting time p, then probability of CU time wasted = p n and, probability of CU time utilized = p n. ere number of process = robability of time waste = (.) ence probability of CU time utilized = [ (.) ] =.. (c) ixed size partition says for each partition only one process should be allocated. So, internal fragmentation: [ ] + [ ] + [ ] + [ ] + [ ] + [ ] + [ ] = = kb. (d) Grant chart for SRT CU scheduling algorithm is: Turn around time = Completion time Arrival time rocess Arrival Time Burst Time Turn Around Time = = = = = Average TAT = Total TAT Number of process = i = = = =. (c) Size of virtual addresses = bits age size = kb age offset = = bits Number of bits used for indexing = = bits Number of set = = Total tag bits = = bits. = = require bits. (c) EMAT = S + ( ) m = age fault rate S = age fault service time
3 Computer Science & IT m = Main memory access time. (a) + + msec = + (.) msec =. +. =. msec Effective access time = T c + ( ) (T m + T c ) where, T c = nsec, T m = nsec and T e = / T c. = + ( ) ( + ) = + = / =.. (b) Option (b) is correct, circular waiting is a necessary condition for deadlock, but not a sufficient condition.. (d) When a user program calls a system call then mode changes to system mode, system call number is used to find out code in system call interrupt vector area and interrupts are enabled.. (c) irst Come irst Serve and Round Robin Scheduling will not have starvation i.e. every process gets CU for its execution. CS will be non preemptive but RR will preempt that process for a while but it is guaranteed that CU time will be given to it.. (b) Mutual exclusion can be solved by spooling everything. old and Wait can be solved by requesting all resources initially. No preemption can be solved by taking request away. Circular wait can be solved by numbering the resources numerically.. (d) rocess Safe sequence Currently Allocated Maximum Need Available A B C A B C A B C rocess Safe sequence Currently Allocated Maximum Need Available A B C A B C A B C
4 CT- CS Operating Systems rocess Currently Allocated Maximum Need Available A B C A B C A B C Safe sequence All options of are safe sequence.. (b) Total seek time = [( )+( ) + ( ) + ( ) + ( ) + ( ) + ( )] msec = [ ] msec = [] msec = msec. (c) Size of disk = GB Disk block size = kb Number of disk blocks in disk = = M blocks In bit-map method each block bit require. So, number of blocks for keeping track of free space M kb = = blocks. (d) Short-term scheduler executes frequently to dispatch the process from the ready queue to CU. So, option (d) is correct and remaining options are the job of long term scheduler.. (b) page faults
5 Computer Science & IT. (a) hysical memory size = MB Number of bits for physical memory = log ( ) = bit. rame size = age size = kb age offset bits = log ( ) = bits x + = x = bits. # rame # age offset x. (c) In fork system call both parent and child will have same virtual address but physical address will be different. When fork == then child process is created and after completion of child process it return its process id to its parent.. (c) bit Average rotational latency = sec =. msec Transfer time = kb / MB =. msec Access time = Seek time + Rotational latency + Transfer time = =. msec. (d) If a process has m resources it can finish and can not be involved in a deadlock. Therefore, the worst case is where every process has m resources and needs another one. If their is one resource left over, one process can finish and replace all its resources, letting the rest to finish. Therefore the condition for avoiding deadlock is r >= (m )p +.. (b). Option (a) is not a valid deadlock prevention scheme.. Option (b) is valid deadlock prevention scheme in which resources are numbered uniquely and resources are assign to process only in increasing order.. Option (c) is deadlock avoidance scheme but not deadlock prevention scheme.. (c) S : or better throughput larger block size used in fixed block size file system but it results decreases in the space utilization. S : Number of DBA s possible in one disk block = DBA Size DBA. (a) (a) If mutual exclusion is satisfied then only one process execute inside the critical section, hence there is no race condition (when two or more process can enter at a same time, then execution order will be different). (b) A process remains inside critical section for only finite amount of time. (c) It is the progress property according to it if no process is executing in its critical section, then selection of a process that will be allowed to enter its critical section can t be postponed for indefinitely..
6 CT- CS Operating Systems. (c) Time quantum = q units Switching time = s units rocess time = r units Number of switches required = r / q So, switching time = s units r / q s units. So total time for switching processes = Total switches Switch time = ( ). (c) age size = KB bit offset Number of frame bits = = bits age table entry = Valid + Translation (frame bits) = + rame bits = + = bits age table size = Mbytes Number of pages = Number of page table entries Mbytes = = M = pages Bits bits needed for page and bits offset Length of virtual address = + = bits.. (b) Contiguous file allocation results in external fragmentation because it could be the case that there many small unallocated fragments between files and they can t be used to allocate a file whose size greater than the size of any fragment but less than total size of free space. Linked allocation supports sequential access to disk block but not random access. Linked allocation does not suffer from external fragmentation. In indexed file allocation, one block is used for each file as an index to store only block pointers.. (b) By using LRU policy: faults By using IO policy: faults So, [LRU IO] page fault: =
7 Computer Science & IT. (c) Data block size Data block size Data block size Total file size = Direct DBA+# + # + # Data block size DBA DBA DBA Data block size = bits Data block size DBA = # Disk block address store inside one block Maximum file size = [ + *() + *() + *() ] bits = k-bits. (c) In given code (a) decrement the value of semaphore and V(a) increment the value of semaphore. X and Y are wait and signal operations on binary semaphore b and c. To avoid mutual exclusion, the following could be done. E: Y(b) E : Y( c) E : X( c) E : Y(b). (a) τ n+ τ τ τ = α T n + ( α) τ n =. +. = Unit =. +. = Unit =. +. =. Unit. (b) age size = kb = bytes age table entry = B = bytes st level page table = # page table entries age table entry size = bytes = = bytes nd level page table = # page table entries age table entry size = bytes = = bytes rd level page table = # page table entries age table entry size = bytes = = bytes Now page size = rd level size. So no need of further level.
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